Several methods are in common use for testing
$H_0: p_1=p_2$ against $H_a: p_l\ne p_2.$
One method (called 'pooled') uses the null hypothesis $p_1=p_2=p$ to estimate
$\hat p = \frac{x_1+x_2}{n_1+n_2} = \frac{48+56}{550+450}
= \frac{104}{1000} = 0.104$ and then base the (estimated) standard
error of $\hat p_1 - \hat p_2$ (denominator of $Z)$
as $\sqrt{\hat p(1-\hat p)(1/n_1+1/n_2)}.$ See NIST handbook.
Another method is to estimate $Var(\hat p_i), i = 1,2$ separately as $\hat\sigma_i =\hat p_i(1-\hat p_i)/n_i,$ respectively.
Then the (estimated) standard error of $\hat p_1 - \hat p_2$ is
$\sqrt{\hat\sigma_1 + \hat\sigma_2}.$ [See, e.g., Ott & Longnecker: Intro. to Stat. Meth. and Data Analysis, Ch 10.] An advantage of this method is that the estimated standard error is also appropriate for making a CI of
$p_1 - p_2.$
Both methods have been used in software implementations of the test for two binomial proportions, sometimes with
various continuity corrections. Minitab software uses
this method as default, but allows the user to specify
the pooled method.
From a recent release of Minitab, here is output from
the 'separate variances' method, with P-value $0.059.$
Test and CI for Two Proportions
Sample X N Sample p
1 48 550 0.087273
2 56 450 0.124444
Difference = p (1) - p (2)
Estimate for difference: -0.0371717
95% CI for difference: (-0.0757267, 0.00138324)
Test for difference = 0 (vs ≠ 0):
Z = -1.89 P-Value = 0.059
Also, using the 'pooled' method, the P-value is $0.055.$
However, the 95% CI for $p_1 - p_2$ uses the separate variances method, so the CI is exactly the same in both outputs.
Test and CI for Two Proportions
Sample X N Sample p
1 48 550 0.087273
2 56 450 0.124444
Difference = p (1) - p (2)
Estimate for difference: -0.0371717
95% CI for difference: (-0.0757267, 0.00138324)
Test for difference = 0 (vs ≠ 0):
Z = -1.92 P-Value = 0.055
Output from the procedure prop.test in R, agrees
with the separate variances version of Minitab except
for differences in rounding.
prop.test(c(48,56), c(550,450), cor=F)
2-sample test for equality of proportions
without continuity correction
data: c(48, 56) out of c(550, 450)
X-squared = 3.6699, df = 1, p-value = 0.0554
alternative hypothesis: two.sided
95 percent confidence interval:
-0.07572667 0.00138324
sample estimates:
prop 1 prop 2
0.08727273 0.12444444
Essentially the same test can be done in R, using a chi-squared test
(chisq.test), which uses for input a $2\times 2$ table of successes and failures.
TBL = rbind(c(48,56), c(550-48, 450-56)); TBL
[,1] [,2]
[1,] 48 56
[2,] 502 394
chisq.test(TBL, cor=F)
Pearson's Chi-squared test
data: TBL
X-squared = 3.6699, df = 1, p-value = 0.0554
On account of the relatively small counts, some statisticians would use Fisher's exact test, which gives P-value $0.061,$ also failing to reject at the 5% level. [See Wikipedia.] This P-value for Fisher's exact test is also
shown in Minitab printouts of the pooled and separate
variances methods, but was omitted above.
fisher.test(TBL)
Fisher's Exact Test for Count Data
data: TBL
p-value = 0.06102
alternative hypothesis:
true odds ratio is not equal to 1
95 percent confidence interval:
0.4375136 1.0319594
sample estimates:
odds ratio
0.6730183
